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4Transfer Functions
• Convenient representation of a linear, dynamic model.
• A transfer function (TF) relates one input and one output:
( )( )
( )( )
systemx t y tX s Y s
→ →
The following terminology is used:
y
output
response
“effect”
x
input
forcing function
“cause”
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4Definition of the transfer function:Let G(s) denote the transfer function between an input, x, and an output, y. Then, by definition
( ) ( )( )
Y sG s
X swhere:
( ) ( )( ) ( )
Y s y t
X s x t
L
L
Development of Transfer Functions
Example: Stirred Tank Heating System
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4
Figure 2.3 Stirred-tank heating process with constant holdup, V.
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4Recall the previous dynamic model, assuming constant liquid holdup and flow rates:
( ) (1)idTV C wC T T Qdt
ρ = − +
Suppose the process is initially at steady state:
( ) ( ) ( ) ( )0 , 0 , 0 2i iT T T T Q Q= = =
where steady-state value of T, etc. For steady-state conditions:
T
( )0 (3)iwC T T Q= − +
Subtract (3) from (1):
( ) ( ) ( ) (4)i idTV C wC T T T T Q Qdt
ρ = − − − + −
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4But,
( )because is a constant (5)
d T TdT Tdt dt
−=
Thus we can substitute into (4-2) to get,
( ) (6)idTV C wC T T Qdt
ρ′
′ ′ ′= − +
where we have introduced the following “deviation variables”, also called “perturbation variables”:
, , (7)i i iT T T T T T Q Q Q′ ′ ′− − −
Take L of (6):
( ) ( ) ( ) ( ) ( )0 (8)iV C sT s T t wC T s T s Q sρ ′ ′ ′ ′ ′ − = = − −
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4( )0 .T t′ =Evaluate
By definition, Thus at time, t = 0,.T T T′ −
( ) ( )0 0 (9)T T T′ = −
But since our assumed initial condition was that the process was initially at steady state, i.e., it follows from (9) that
Note: The advantage of using deviation variables is that the initial condition term becomes zero. This simplifies the later analysis.
( )0T T=( )0 0.T ′ =
Rearrange (8) to solve for T s( ) :′
( ) ( ) ( )1 (10)1 1 i
KT s Q s T ss sτ τ
′ ′ ′= + + +
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4where two new symbols are defined:
( )1 and 11VKwC w
ρτ
Transfer Function Between and Q′ T ′
Suppose is constant at the steady-state value. Then, iT
Then we can substitute into (10) and rearrange to get the desired TF: ( ) ( ) ( )0 0.i i i iT t T T t T s′ ′= ⇒ = ⇒ =
( )( )
(12)1
T s KQ s sτ′
=′ +
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4Transfer Function Between and T ′ :iT ′
Suppose that Q is constant at its steady-state value:
( ) ( ) ( )0 0Q t Q Q t Q s′ ′= ⇒ = ⇒ =
Thus, rearranging
( )( )
1 (13)1i
T sT s sτ′
=′ +
Comments:
1. The TFs in (12) and (13) show the individual effects of Q and on T. What about simultaneous changes in both Q and ?iT
iT
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4• Answer: See (10). The same TFs are valid for simultaneous
changes.
• Note that (10) shows that the effects of changes in both Qand are additive. This always occurs for linear, dynamic models (like TFs) because the Principle of Superposition is valid.
iT
2. The TF model enables us to determine the output response to any change in an input.
3. Use deviation variables to eliminate initial conditions for TF models.
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4Properties of Transfer Function Models
1. Steady-State Gain
The steady-state of a TF can be used to calculate the steady-state change in an output due to a steady-state change in the input. For example, suppose we know two steady states for an input, u, and an output, y. Then we can calculate the steady-state gain, K, from:
2 1
2 1(4-38)y yK
u u−
=−
For a linear system, K is a constant. But for a nonlinear system, K will depend on the operating condition ( ), .u y
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4Calculation of K from the TF Model:
If a TF model has a steady-state gain, then:
( )0
lim (14)s
K G s→
=
• This important result is a consequence of the Final Value Theorem
• Note: Some TF models do not have a steady-state gain (e.g., integrating process in Ch. 5)
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42. Order of a TF Model
Consider a general n-th order, linear ODE:1
1 1 01
1
1 1 01 (4-39)
n n m
n n mn n m
m
m m
d y dy dy d ua a a a y bdtdt dt dt
d u dub b b udtdt
−
− −
−
− −
+ + + = +
+ + +
…
…
Take L, assuming the initial conditions are all zero. Rearranging gives the TF:
( ) ( )( )
0
0
(4-40)
mi
iin
ii
i
b sY s
G sU s
a s
=
=
= =∑
∑
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4Definition:
The order of the TF is defined to be the order of the denominator polynomial.
Note: The order of the TF is equal to the order of the ODE.
Physical Realizability:For any physical system, in (4-38). Otherwise, the system response to a step input will be an impulse. This can’t happen.
Example:
n m≥
0 1 0 and step change in (4-41)dua y b b u udt
= +
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43. Additive Property
Suppose that an output is influenced by two inputs and that the transfer functions are known:
( )( ) ( ) ( )
( ) ( )1 21 2
andY s Y s
G s G sU s U s
= =
Then the response to changes in both and can be written as:
1U 2U
( ) ( ) ( ) ( ) ( )1 1 2 2Y s G s U s G s U s= +
The graphical representation (or block diagram) is:
G1(s)
G2(s)
Y(s)U1(s)
U2(s)
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44. Multiplicative Property
Suppose that,
( )( ) ( ) ( )
( ) ( )22 3
2 3and
Y s U sG s G s
U s U s= =
Then,( ) ( ) ( ) ( ) ( ) ( )2 2 2 3 3Y s G s U s and U s G s U s= =
Substitute,
( ) ( ) ( ) ( )2 3 3Y s G s G s U s=
Or,
( )( ) ( ) ( ) ( ) ( ) ( ) ( )2 3 3 2 3
3
Y sG s G s U s G s G s Y s
U s=
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4Linearization of Nonlinear Models• So far, we have emphasized linear models which can be
transformed into TF models.
• But most physical processes and physical models are nonlinear.
- But over a small range of operating conditions, the behavior may be approximately linear.
- Conclude: Linear approximations can be useful, especially for purpose of analysis.
• Approximate linear models can be obtained analytically by a method called “linearization”. It is based on a Taylor Series Expansion of a nonlinear function about a specified operating point.
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4• Consider a nonlinear, dynamic model relating two process
variables, u and y:
( ), (4-60)dy f y udt
=
Perform a Taylor Series Expansion about and and truncate after the first order terms,
u u= y y=
( ) ( ), , (4-61)y y
f ff u y f u y u yu y∂ ∂′ ′= + +∂ ∂
where and . Note that the partial derivative terms are actually constants because they have been evaluated atthe nominal operating point,
Substitute (4-61) into (4-60) gives:
u u u′ = − y y y′ = −
( ), .u y
( ),y y
dy f ff u y u ydt u y
∂ ∂′ ′= + +∂ ∂
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4The steady-state version of (4-60) is:
( )0 ,f u y=
,dy dydt dt
′=Substitute into (7) and recall that
(4-62)y y
dy f fu ydt u y′ ∂ ∂′ ′= +
∂ ∂Linearizedmodel
Example: Liquid Storage System
Mass balance:
Valve relation:
A = area, Cv = constant
(1)idhA q qdt
= −
(2)vq C h=
qi
h
q
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4Combine (1) and (2),
(3)i vdhA q C hdt
= −
Linearize term,
( )1 (4)2
h h h hh
≈ − −
Or
1 (5)h h hR
′≈ −
where:2R h
h h h′ −
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4Substitute linearized expression (5) into (3):
1 (6)i vdhA q C h hdt R
′= − −
The steady-state version of (3) is:
0 (7)i vq C h= −
Subtract (7) from (6) and let , noting that gives the linearized model:
i i iq q q′ − dh dhdt dt
′=
1 (8)idhA q hdt R′
′ ′= −
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4Summary:
In order to linearize a nonlinear, dynamic model:
1. Perform a Taylor Series Expansion of each nonlinear term and truncate after the first-order terms.
2. Subtract the steady-state version of the equation.
3. Introduce deviation variables.
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4State-Space Models
• Dynamic models derived from physical principles typically consist of one or more ordinary differential equations (ODEs).
• In this section, we consider a general class of ODE models referred to as state-space models.
Consider standard form for a linear state-space model,
(4-90)
(4-91)
x = Ax + Bu + Ed
y = Cx
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4where:
x = the state vector
u = the control vector of manipulated variables (also called control variables)
d = the disturbance vector
y = the output vector of measured variables. (We use boldface symbols to denote vector and matrices, and plain text to represent scalars.)
• The elements of x are referred to as state variables.
• The elements of y are typically a subset of x, namely, the state variables that are measured. In general, x, u, d, and y are functions of time.
• The time derivative of x is denoted by
• Matrices A, B, C, and E are constant matrices.
( )d / d .t=x x
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4Example 4.9Show that the linearized CSTR model of Example 4.8 canbe written in the state-space form of Eqs. 4-90 and 4-91.Derive state-space models for two cases:
(a) Both cA and T are measured.
(b) Only T is measured.
Solution
The linearized CSTR model in Eqs. 4-84 and 4-85 can be written in vector-matrix form:
11 12
21 22 2
0(4-92)
A A
s
dc a a cdt T
dTa a T b
dt
′ ′ ′= + ′ ′
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4Let and , and denote their time derivatives by and . Suppose that the steam temperature Ts can be manipulated. For this situation, there is a scalar control variable,
, and no modeled disturbance. Substituting these definitions into (4-92) gives,
1 Ax c′ 2x T ′ 1x2x
su T ′
1 11 12 1
2 21 22 2 2
0(4-93)
x a a xu
x a a x b
= +
BA
which is in the form of Eq. 4-90 with x = col [x1, x2]. (The symbol “col” denotes a column vector.)
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4a) If both T and cA are measured, then y = x, and C = I in
Eq. 4-91, where I denotes the 2x2 identity matrix. A and B are defined in (4-93).
b) When only T is measured, output vector y is a scalar, and C is a row vector, C = [0,1].y T ′=
Note that the state-space model for Example 4.9 has d = 0because disturbance variables were not included in (4-92). By contrast, suppose that the feed composition and feed temperatureare considered to be disturbance variables in the original nonlinear CSTR model in Eqs. 2-60 and 2-64. Then the linearizedmodel would include two additional deviation variables, and .
Aic′iT ′
